What is the derivative of \sqrt[4]{x^3+4x^2+7} ?

The derivative of \sqrt[4]{x^3+4x^2+7} is (3x^2+8x)/(4(x^3+4x^2+7)^{3/4)}

What is the first derivative of \sqrt[4]{x^3+4x^2+7} ?

The first derivative of \sqrt[4]{x^3+4x^2+7} is (3x^2+8x)/(4(x^3+4x^2+7)^{3/4)}

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\frac{3 x ^{2} + 8 x}{4 ( x ^{3} + 4 x ^{2} + 7 ) ^{\frac{3}{4}}}

Solution steps

Apply the chain rule:

\frac{1}{4 ( x ^{3} + 4 x ^{2} + 7 ) ^{\frac{3}{4}}} \frac{d}{d x} (x^{3} + 4 x^{2} + 7)

= \frac{1}{4 ( x ^{3} + 4 x ^{2} + 7 ) ^{\frac{3}{4}}} \frac{d}{d x} (x^{3} + 4 x^{2} + 7)

\frac{d}{d x} (x^{3} + 4 x^{2} + 7) = 3 x^{2} + 8 x

= \frac{1}{4 ( x ^{3} + 4 x ^{2} + 7 ) ^{\frac{3}{4}}} (3 x^{2} + 8 x)

Simplify

\frac{1}{4 ( x ^{3} + 4 x ^{2} + 7 ) ^{\frac{3}{4}}} (3 x^{2} + 8 x) : \frac{3 x ^{2} + 8 x}{4 ( x ^{3} + 4 x ^{2} + 7 ) ^{\frac{3}{4}}}

= \frac{3 x ^{2} + 8 x}{4 ( x ^{3} + 4 x ^{2} + 7 ) ^{\frac{3}{4}}}

\frac{d}{d x} (3 x \cdot sin (x))

\frac{( x )}{4 + x}

0.49489 n v^{2} + 1500 \cdot \frac{n}{v}

\int 3 x^{2} + 8 x d x

\frac{\partial}{\partial x} (6 x y^{2})

What is the derivative of \sqrt[4]{x^3+4x^2+7} ?
The derivative of \sqrt[4]{x^3+4x^2+7} is (3x^2+8x)/(4(x^3+4x^2+7)^{3/4)}
What is the first derivative of \sqrt[4]{x^3+4x^2+7} ?
The first derivative of \sqrt[4]{x^3+4x^2+7} is (3x^2+8x)/(4(x^3+4x^2+7)^{3/4)}